Question

A ball is projected from the ground with velocity $v$ such that its range is maximum
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{i. Velocity at half of the maximum height in vertical direction}& a.\frac{v}{2}\\ \text{ii Velocity at the maximum height}& b.\frac{v}{\sqrt{2}}\\ \text{iii. Change in its velocity when it returns}& c.v\sqrt{2}\\ \text{to the ground}& \\ \text{iv. Average velocity when it reaches}& d.\frac{v}{2}\sqrt{\frac{5}{2}}\\ \text{the maximum height}& \end{array}$

• A. i- a ii-b iii- d iv- c
• B. i- a ii-b iii- c iv- d
• C. i- b ii-a iii- d iv- c
• D. i- a ii-d iii- b iv- c

Answer 1

The correct option is B i- a ii-b iii- c iv- d

$i.\to a.,ii.\to b.,iiii.\to c.,iv.\to d.$
$i.\to a$.
Range is maximum, when the angle of projection is ${45}^{\circ }$
$H=\frac{v^2}{2g}si{n}^2{45}^{\circ };H=\frac{v^2}{4g}$
Velocity, at half of the maximum height is $v^{\prime }$
$v^{\prime 2}=v^{2}si{n}^2{45}^{\circ }-2g\frac{H}{2};v^{\prime 2}=\frac{v^2}{2}-\frac{v^2}{4}$
$v^{\prime 2}=\frac{v^2}{4}\Rightarrow v^{\prime }=\frac{v}{2}$
$ii\to b$.
Velocity at the maximum height
$v^{\prime }=\text{v cos}{45}^{\circ }\Rightarrow v^{\prime }=\frac{v}{\sqrt{2}}$
[because vertical component of velocity is zero at the highest point]
$iii\to c.$
Projection velocity
At projection point, $v_i=\text{v cos}{45}^{\circ }\hat{i}+\text{v sin}{45}^{\circ }\hat{j}$
At the point , when the body strikes the ground
$\overrightarrow{v_f}=v\text{cos}{45}^{\circ }\hat{i}-v\text{sin}{45}^{\circ }\hat{j}$
$\Delta v=\overrightarrow{v_f}+(-\overrightarrow{v_f})=2\text{v sin}{45}^{\circ }(-j)$
$|\Delta \overrightarrow{v}|=2\text{v sin}{45}^{\circ }=v\sqrt{2}$

$iv\to d.\text{Average velocity}=\frac{\text{Total displacement}}{\text{Total time}}$

$\text{Displacement}=\sqrt{{\left(\frac{R}{2}\right)}^2+H^2}$
${\overrightarrow{v}}_{av}=\frac{\sqrt{\frac{R^2}{4}+H^2}}{\frac{v\sin \theta }{g}}=\frac{\sqrt{R^2+4H^2}}{\frac{2v\sin \theta }{g}}$
$v_{av}=\frac{{\left(\frac{v^2}{g}\right)}^2+4{\left(\frac{v^2}{4g}\right)}^2}{\frac{\sqrt{2}v}{g}}$
$=\frac{\frac{v^2}{g}\sqrt{1+\frac{1}{4}}}{\frac{v\sqrt{2}}{g}}=\frac{v^2\sqrt{5}}{2v\sqrt{2}}=\frac{v}{2}\sqrt{\frac{5}{2}}$